Titration of 20.0 mL 0.120 M HCHO2 with 0.0800 M NaOH. Find the pH: (a) before NaOH addition; (b) after addition of 15.0 mL NaOH; (c) after addition of 30.0 mL NaOH; (d) after addition of 30.5 mL NaOH.

Explanation:

Given:

Ka = 1.8 × 10-4

HCOOH --> HCOO- + H+

HA --> H+ + A-

Ka = [H+] × [A-]/[HA]

[HA] = 0.12 M

[H+] = [A-] = x^2

1.8 × 10^-4 × 0.12 = x^2

x = 0.00465 M

pH = - log [H+]

= -log [0.00465]

= 2.33

B.

Since 1 mole of NaOH reactes with 1 mole of HCOOH.

Number of moles of NaOH = 0.08 × 15 × 10^-3

= 0.0012 mole.

Concentration of HCOOH = 0.0012/20 × 10^-3

= 0.06 M

[HA] = 0.06 M

1.8 × 10^-4 × 0.06 = x^2

x = 0.00329 M

pH = - log [H+]

= -log [0.00329]

= 2.48

C.

Since 1 mole of NaOH reacts with 1 mole of HCOOH.

Number of moles of NaOH = 0.08 × 30 × 10^-3

= 0.0024 mole.

Concentration of HCOOH = 0.0024/20 × 10^-3

= 0.12 M

[HA] = 0.12 M

1.8 × 10^-4 × 0.12 = x^2

x = 0.00465 M

pH = - log [H+]

= -log [0.00465]

= 2.33

D.

Since 1 mole of NaOH reactes with 1 mole of HCOOH.

Number of moles of NaOH = 0.08 × 30.5 × 10^-3

= 0.00244 mole.

Concentration of HCOOH = 0.00244/20 × 10^-3

= 0.122 M

[HA] = 0.122 M

1.8 × 10^-4 × 0.122 = x^2

x = 0.00469 M

pH = - log [H+]

= -log [0.00469]

= 2.33

the dependant variable is spilled water

the independent is the heat

fe 55.85 g/mol and h2o 18.02 g/mol.